Question

Circuit Analysis In Exercises 73 and 74, consider the circuit shown in the figure. The currents \(I_{1}, I_{2}\), and \(I_{3}\), in amperes, are the solution of the system of linear equations \(\left\\{\begin{aligned} 2 I_{1} \quad+4 I_{3}=& E_{1} \\ I_{2}+4 I_{3}=& E_{2} \\ I_{1}+I_{2}-I_{3}=& 0 \end{aligned}\right.\) where \(E_{1}\) and \(E_{2}\) are voltages. Use the inverse of the coefficient matrix of this system to find the unknown currents for the given voltages. \(E_{1}=28\) volts, \(E_{2}=21\) volts

Short answer

The currents flowing through the circuit are \(I_{1} = -7 \, A\), \(I_{2} = 3 \, A\), and \( I_{3} = 2.5 \, A\)

Step-by-step solution

Formulating the Coefficient and Constant Matrices

We have a system of equations in the form \(AX=B\), where \(A\) is the coefficient matrix, \(X\) is the current matrix, and \(B\) is the constant matrix. We can write the system as \(\left[ \begin{array} {ccc} 2 & 0 & 4 \ 0 & 1 & 4 \ 1 & 1 & -1 \end{array} \right] \left[ \begin{array} {c} I_{1} \ I_{2} \ I_{3} \end{array} \right] = \left[ \begin{array} {c} E_{1} \ E_{2} \ 0 \end{array} \right]\). So, \(A = \left[ \begin{array} {ccc} 2 & 0 & 4 \ 0 & 1 & 4 \ 1 & 1 & -1 \end{array} \right]\), \(X = \left[ \begin{array} {c} I_{1} \ I_{2} \ I_{3} \end{array} \right]\), and \(B = \left[ \begin{array} {c} E_{1} \ E_{2} \ 0 \end{array} \right]\).

Calculating the Inverse of the Coefficient Matrix

We can use the formula for 3x3 matrices to get the inverse of the matrix A: \(A^{-1} = \frac{1}{det(A)} adj(A)\), where \(adj(A)\) is the adjugate of \(A\), and \(det(A)\) is the determinant of \(A\). We first calculate the determinant, which is \(2=(-2)*(-1)-0-4*(-4)=8\). Then we get the adjugate by taking the transpose of the cofactor matrix. The resultant inverse matrix is \(A^{-1}=\frac{1}{8}\left[ \begin{array} {ccc} -4 & -8 & 4 \ 4 & 2 & 0 \ 0 & 2 & 2 \end{array} \right]\).

Determining the Unknown Currents

We can now find the unknown currents by multiplying \(A^{-1}\) and \(B\). It leads to the solution: \(X = A^{-1}B\) , substituting the given \(E_{1}=28\) volts, \(E_{2}=21\) volts to \(B\) matrix, we get \(X = \frac{1}{8}\left[ \begin{array} {ccc} -4 & -8 & 4 \ 4 & 2 & 0 \ 0 & 2 & 2 \end{array} \right] \left[ \begin{array} {c} 28 \ 21 \ 0 \end{array} \right], = \left[ \begin{array} {c} -7 \ 3 \ 2.5 \end{array} \right]\) So, \(I_{1} = -7 \, A\), \(I_{2} = 3 \, A\), and \( I_{3} = 2.5 \, A\). These are the currents flowing through the circuit.

Key concepts

System of Linear Equations

A system of linear equations consists of multiple linear equations that are solved simultaneously. Each equation relates several variables with coefficients, often representing real-world relationships, such as electrical currents in a circuit. In our example, currents \(I_{1}\), \(I_{2}\), and \(I_{3}\) are determined by the system that includes the resistance and voltage values in a circuit. The solution to the system provides the values of these currents that satisfy all the equations at once.

To solve such systems, one can use elimination, substitution, graphical methods, or matrix operations such as finding the inverse of a matrix. Using matrix operations is particularly efficient for larger systems or ones that are not easily simplified by hand, which is why they are often preferred in engineering and physics problems like circuit analysis.

Inverse of a Matrix

The inverse of a matrix \(A\) is another matrix, denoted as \(A^{-1}\), which when multiplied with \(A\) yields the identity matrix. The identity matrix is the matrix equivalent of the number 1 in arithmetic; it leaves any matrix unchanged when multiplied by it. However, not all matrices have an inverse; a matrix must be square (same number of rows and columns) and its determinant (a scalar value) cannot be zero.

Calculating the inverse involves finding the determinant and the adjugate (transpose of the cofactor matrix) of \(A\). This is feasible for 2x2 and 3x3 matrices by hand, but for larger matrices, it's more practical to use computational tools. The inverse is particularly useful for solving a system of equations represented in matrix form \(AX = B\) because it helps to find \(X\) by computing \(X = A^{-1}B\), thus avoiding the need for more complex algebraic manipulations.

Coefficient Matrix

In the context of systems of linear equations, the coefficient matrix is a square matrix that contains only the coefficients of the variables in the system. It's an organized representation that expresses the data neatly when coupled with matrix operations.

In our exercise, the coefficient matrix \(A\) is composed of the coefficients of currents \(I_{1}\), \(I_{2}\), and \(I_{3}\) from the three equations of the circuit. The matrix does not include the voltage values, which are instead part of the constant matrix \(B\). Upon finding the inverse of the coefficient matrix, it's possible to determine the values of the currents by calculating the product of \(A^{-1}\) with matrix \(B\), which simplifies the solution process enormously as compared to solving the system equation by equation.

Electrical Currents Calculation

Solving for electrical currents in a circuit using matrix operations simplifies and speeds up the process, especially for complex circuits. Once the currents have established relationships via Kirchhoff's laws and Ohm's law, these currents can be represented in a system of linear equations. With matrices, the physical rules governing the circuit become solvable through algebra, and one can easily calculate the currents flowing through the circuit components.

As we've seen in the provided solution, once the inverse of the coefficient matrix is obtained, it's multiplied by the constant matrix (voltage values in our case) to find the matrix of the currents (\(\mathbf{X}\)). The resulting vector gives the exact values of the currents, allowing engineers and physicists to analyze and make informed decisions about the circuit's behavior and safety.